Figure 14.11 shows a gravity retaining wall for a granular (c′ = 0) backfill. The same soil is present at the bottom of the wall and on the left. The unit weight and the friction angle of the backfill are 18.5 kN/m³ and 35°, respectively. The unit weight of the concrete is 24.0 kN/m³. Determine

Question

Figure 14.11 shows a gravity retaining wall for a granular (c′ = 0) backfill. The same soil is present at the bottom of the wall and on the left. The unit weight and the friction angle of the backfill are 18.5 kN/m³ and 35°, respectively. The unit weight of the concrete is 24.0 kN/m³. Determine the factors of safety with respect to overturning, sliding, and bearing capacity failure.

Accepted Answer

The retaining wall and the soil regions of interest are divided into the rectangles and triangles shown in Figure 14.12 for case of computation of the weights and moment arms.
For φ′ = 35° and α = 15°, we find from Table 11.2 that [latex]K_a[/latex] = 0.2968. Thus,
[latex]P_a=\frac{1}{2}\gamma H^{\prime 2}K_a=\frac{1}{2}(18.5)(6.54)^2(0.2968)=[/latex] 117.4 kN/m
[latex]P_h=P_a \cos \alpha =[/latex] 113.4 kN/m
[latex]P_v=P_a \sin \alpha=[/latex] 30.4 kN/m

Section	Weight (kN/m)	Moment arm from C (m)	Moment about C (kN⋅m/m)
1	[latex](0.5)(1.5)(6.0)(24.0)=108.0[/latex]	1.0	108.0
2	[latex](0.5)(6.0)(24.0)=72.0[/latex]	1.75	126.0
3	[latex](0.5)(2.0)(6.0)(24.0)=144.0[/latex]	2.67	384.5
4	[latex](0.5)(2.0)(6.0)(18.5)=111.0[/latex]	3.33	369.6
5	[latex](0.5)(2.0)(0.54)(18.5)=10.0[/latex]	3.33	33.3
	[latex]P_v=30.4[/latex]	4.0	121.6
	[latex]\sum V=475.4[/latex]		[latex]\sum M_R=1143.0[/latex]

Stability with Respect to Overturning
The overturning moment is
[latex]M_O=P_h\frac{H^\prime}{3}=113.4 imes 2.18=[/latex] 247.2 kN/m
The resisting moment is
[latex]\sum M_R =[/latex] 1143.0 kN/m
Therefore,
[latex]FS_{(overturning)} = \frac{\sum M_g}{M_O}=\frac{1143.0}{247.2} =[/latex] 4.62
Stability with Respect to Sliding
For φ′ = 35°, from Eq. 11.28,
[latex]K_p= an^2\left(45+\frac{\phi^\prime}{2} ight)=[/latex] 3.690
[latex]P_p=\frac{1}{2}\gamma D^2 K_p=\frac{1}{2}(18.5)(1.0)^2(3.690)=[/latex] 34.1 kN/m
Take δ′ as [latex]{}^2/3[/latex] φ′. So δ′ = 23.3°, we have
[latex]FS_{(sliding)} =\frac{ (\sum V) an \delta^\prime+ P_p}{P_h}=\frac{475.4 imes an 23.3 + 34.1}{113.4} =[/latex] 2.11
Stability with Respect to Bearing Capacity Failure
[latex]e = \frac{B}{2}-\frac{ \sum M_R - \sum M_O}{\sum V} =\frac{ 4.0}{2} -\frac{ 1143.0 - 247.2}{475.4} = 0.116 m \lt \frac{B}{6} \left(=\frac{4}{6}=0.67 m ight)[/latex]
From Eq. (14.11),
[latex]q_{max} = q_{toe} = \frac{\sum V}{B}\left(1+\frac{6e}{B} ight)=\frac{475.4}{4.0} \left(1+\frac{(6)(0.116)}{4.0} ight)=[/latex] 139.5 kN/m²
In granular soil, Eq. 14.13 becomes
[latex]q_u^\prime = q N_qF_{qd} F_{qi} +0.5 \gamma B^\prime N_\gamma F_{\gamma d}F_{\gamma i}[/latex]
where, B′ = B - 2e = 3.768 m and q = γD = (18.5)(1.0) = 18.5 kN/m².
For φ′ = 35°, from Table 12.1

φ′	[latex]N_c[/latex]	[latex]N_q[/latex]	[latex]N_\gamma[/latex]	φ′	[latex]N_c[/latex]	[latex]N_q[/latex]	[latex]N_\gamma[/latex]
0	5.14	1.00	0.00	23	18.05	8.66	8.20
1	5.38	1.09	0.07	24	19.32	9.60	9.44
2	5.63	1.20	0.15	25	20.72	10.66	10.88
3	5.90	1.31	0.24	26	22.25	11.85	12.54
4	6.19	1.43	0.34	27	23.94	13.20	14.47
5	6.49	1.57	0.45	28	25.80	14.72	16.72
6	6.81	1.72	0.57	29	27.86	16.44	19.34
7	7.16	1.88	0.71	30	30.14	18.40	22.40
8	7.53	2.06	0.86	31	32.67	20.63	25.99
9	7.92	2.25	1.03	32	35.49	23.18	30.22
10	8.35	2.47	1.22	33	38.64	26.09	35.19
11	8.80	2.71	1.44	34	42.16	29.44	41.06
12	9.28	2.97	1.69	35	46.12	33.30	48.03
13	9.81	3.26	1.97	36	50.59	37.75	56.31
14	10.37	3.59	2.29	37	55.63	42.92	66.19
15	10.98	3.94	2.65	38	61.35	48.93	78.03
16	11.63	4.34	3.06	39	67.87	55.96	92.25
17	12.34	4.77	3.53	40	75.31	64.20	109.41
18	13.10	5.26	4.07	41	83.86	73.90	130.22
19	13.93	5.80	4.68	42	93.71	85.38	155.55
20	14.83	6.40	5.39	43	105.11	99.02	186.54
21	15.82	7.07	6.20	44	118.37	115.31	224.64
22	16.88	7.82	7.13	45	133.88	134.88	271.76

[latex]N_q = 33.30 [/latex] and [latex]N_\gamma = 48.03[/latex]. Hence,
[latex]F_{qd}=1+2 an\phi^\prime(1-\sin \phi^\prime)^2\frac{D_f}{B}=1+2 an 35(1- \sin 35)^2 \frac{1}{4}=[/latex] 1.07
[latex]F_{\gamma d} =[/latex]1
[latex]\Psi= an^{-1}\left(\frac{P_a \cos \alpha}{\sum V} ight)= an^{-1}\left(\frac{113.4}{475.4} ight)=[/latex]13.4°
[latex]F_{qi}=\left(1-\frac{\Psi}{90} ight)^2=\left(1-\frac{13.4}{90} ight)=[/latex] 0.72
[latex]F_{\gamma i}=\left(1+\frac{\Psi}{\phi^\prime} ight)^2=\left(1-\frac{15}{35} ight)^2=[/latex] 0.33
[latex]q_u = (18.5)(33.30)(1.07)(0.72) + (0.5)(18.5)(3.768)(48.03)(1.0)(0.33)=[/latex] 1027.0 kN/m²
[latex]FS_{(bearing capacity)} =\frac{q_u^\prime}{q_{max}}=\frac{1027.0}{139.5} =[/latex] 7.4

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